LeetCode 2544 - Alternating Digit Sum

The problem asks us to compute an alternating sum of the digits of a positive integer. The alternation starts from the most significant digit, which always has a positive sign. Every following digit flips the sign from the previous one.

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LeetCode Problem 2544

Difficulty: 🟢 Easy
Topics: Math

Solution

Problem Understanding

The problem asks us to compute an alternating sum of the digits of a positive integer. The alternation starts from the most significant digit, which always has a positive sign. Every following digit flips the sign from the previous one.

For example, if the number is 521, the calculation becomes:

  • +5
  • -2
  • +1

The final answer is 4.

The input is a single positive integer n. We need to examine each digit in order from left to right and apply alternating signs. The output is a single integer representing the signed sum.

The constraints are small:

1 <= n <= 10^9

This means the number contains at most 10 digits. Even a straightforward digit-by-digit solution is extremely fast. Because the input size is tiny, efficiency is not a concern in practice, but we should still design a clean and correct algorithm.

One important detail is that the sign assignment depends on the position of the digit from the left side, not the right side. This matters because extracting digits using % 10 naturally processes digits from right to left. A naive implementation that alternates signs while processing from the least significant digit could produce incorrect results unless it carefully compensates for direction.

The problem guarantees that n is positive, so we never need to handle negative numbers or empty input.

Some important edge cases include:

  • A single-digit number, where the answer is simply the digit itself.
  • Numbers with an even number of digits, because the last sign becomes negative.
  • Numbers with repeated digits, where it is easy to accidentally lose track of sign alternation.
  • Numbers ending in zero, where digit extraction still needs to preserve the correct alternating pattern.

Approaches

Brute Force Approach

A brute-force solution converts the integer into a string so we can process digits from left to right directly. We iterate through each character, convert it back to an integer digit, and alternate the sign based on the index.

For index 0, the digit is added.

For index 1, the digit is subtracted.

For index 2, the digit is added again, and so on.

This works because the string representation preserves the natural left-to-right digit order required by the problem.

Although this approach is already fast enough for the constraints, it relies on string conversion instead of pure arithmetic operations.

Optimal Approach

A cleaner mathematical solution still processes digits from left to right, but does so efficiently using string traversal or careful arithmetic.

The key observation is that the sign depends entirely on the digit position:

  • Even index positions contribute positively
  • Odd index positions contribute negatively

Because the number has very few digits, we can simply iterate once through all digits and maintain a running total.

This gives a straightforward linear solution with constant extra space.

Approach Time Complexity Space Complexity Notes
Brute Force O(d) O(d) Converts number to string and stores digits
Optimal O(d) O(1) Single pass through digits with alternating signs

Here, d represents the number of digits in n.

Algorithm Walkthrough

Optimal Algorithm

  1. Convert the integer n into its string representation so the digits can be processed from left to right.
  2. Initialize a variable called result to store the running alternating sum.
  3. Iterate through each digit using its index.
  4. Convert the current character into an integer digit.
  5. If the index is even, add the digit to result.

This works because the first digit always has a positive sign, and every second digit after that alternates back to positive. 6. If the index is odd, subtract the digit from result.

Odd-indexed digits always receive a negative sign. 7. After processing all digits, return result.

Why it works

The algorithm directly follows the definition given in the problem statement. The first digit is always added, and every following digit alternates between subtraction and addition based on its position. Since every digit is processed exactly once in left-to-right order, each digit receives the correct sign, guaranteeing the final sum is correct.

Python Solution

class Solution:
    def alternateDigitSum(self, n: int) -> int:
        result = 0
        digits = str(n)

        for index, ch in enumerate(digits):
            digit = int(ch)

            if index % 2 == 0:
                result += digit
            else:
                result -= digit

        return result

The implementation begins by converting the integer into a string so that digits can be processed in the correct left-to-right order.

The variable result stores the running alternating sum. The loop uses enumerate so we can access both the digit and its position. The parity of the index determines whether the digit should be added or subtracted.

Even indices correspond to positive signs, while odd indices correspond to negative signs. After all digits are processed, the final accumulated value is returned.

Go Solution

func alternateDigitSum(n int) int {
    digits := []rune(string(runeSlice(n)))
    result := 0

    for i, ch := range digits {
        digit := int(ch - '0')

        if i%2 == 0 {
            result += digit
        } else {
            result -= digit
        }
    }

    return result
}

func runeSlice(n int) []rune {
    return []rune(fmt.Sprintf("%d", n))
}

A cleaner LeetCode-ready Go implementation is shown below:

import "strconv"

func alternateDigitSum(n int) int {
    digits := strconv.Itoa(n)
    result := 0

    for i, ch := range digits {
        digit := int(ch - '0')

        if i%2 == 0 {
            result += digit
        } else {
            result -= digit
        }
    }

    return result
}

The Go implementation follows the same logic as the Python version. The integer is converted into a string using strconv.Itoa. Each character is converted into its numeric value by subtracting '0'.

Unlike Python, Go requires explicit imports and explicit type conversion from rune values to integers.

Worked Examples

Example 1

Input:

n = 521

Digits processed from left to right:

Index Digit Operation Result
0 5 +5 5
1 2 -2 3
2 1 +1 4

Final answer:

4

Example 2

Input:

n = 111
Index Digit Operation Result
0 1 +1 1
1 1 -1 0
2 1 +1 1

Final answer:

1

Example 3

Input:

n = 886996
Index Digit Operation Result
0 8 +8 8
1 8 -8 0
2 6 +6 6
3 9 -9 -3
4 9 +9 6
5 6 -6 0

Final answer:

0

Complexity Analysis

Measure Complexity Explanation
Time O(d) Each digit is processed exactly once
Space O(1) Only a few variables are used

The algorithm scans through the digits a single time, so the running time grows linearly with the number of digits. Since the maximum number of digits is very small, the solution is extremely efficient.

The extra space usage is constant because we only maintain a running sum and loop variables. Although string conversion internally allocates memory, the auxiliary algorithmic space remains constant relative to the problem constraints.

Test Cases

solution = Solution()

assert solution.alternateDigitSum(521) == 4      # basic alternating example
assert solution.alternateDigitSum(111) == 1      # repeated digits
assert solution.alternateDigitSum(886996) == 0   # result becomes zero

assert solution.alternateDigitSum(5) == 5        # single digit
assert solution.alternateDigitSum(10) == 1       # includes zero
assert solution.alternateDigitSum(21) == 1       # two digits
assert solution.alternateDigitSum(1234) == -2    # even number of digits
assert solution.alternateDigitSum(999999999) == 9  # large repeated digits
assert solution.alternateDigitSum(1000000000) == 1 # upper constraint boundary
assert solution.alternateDigitSum(10101) == 1    # alternating zeros
Test Why
521 Verifies standard alternating behavior
111 Verifies repeated digits
886996 Verifies cancellation to zero
5 Tests single-digit input
10 Tests handling of zero digit
21 Tests smallest multi-digit case
1234 Tests even-length number
999999999 Tests large repeated digits
1000000000 Tests upper constraint boundary
10101 Tests zeros inside the number

Edge Cases

Single-Digit Numbers

A single-digit number is the smallest valid input. Since the most significant digit always has a positive sign, the answer should simply equal the digit itself.

For example:

n = 7

The result is:

+7 = 7

The implementation handles this naturally because the loop runs once, and index 0 is treated as positive.

Numbers Containing Zero

Digits with value 0 can sometimes cause logic mistakes if the implementation skips them or mishandles sign alternation.

For example:

n = 1010

The computation is:

(+1) + (-0) + (+1) + (-0)

The algorithm processes zeros exactly like any other digit, so the alternating pattern remains correct.

Even Number of Digits

Numbers with an even number of digits end with a negative contribution. This can expose off-by-one mistakes in sign handling.

For example:

n = 1234

The calculation becomes:

(+1) + (-2) + (+3) + (-4) = -2

The implementation correctly uses index parity, ensuring the sign alternates properly regardless of the total number of digits.